8
\$\begingroup\$

I'm a student in electrical engineering in Belgium and I use a multiple feedback bandpass filter in one of my projects.

bandpass filter

I tried to find out the transfer function of this filter and this is what I found:

$$ \frac{V_o}{V_i}=-\left( j\omega C+ \frac{2}{R_2}+\frac{R_1+R_3}{R_1R_2R_3}\cdot\frac{1}{j\omega C} \right) ^{-1} $$

With this expression I made a Bode diagram:

bode diagram

Is that normal that the frequencies other than 1kHz are still amplified (not as much but still) ? Is my transfer function right ?

Thanks !

\$\endgroup\$
2
  • 1
    \$\begingroup\$ It seems a normal transfer function for a bandpass filter...you can't expect a rectangle, and you have -20 dB for a factor 2 frequency multiplication-division...it's compatible with a 2nd order filter \$\endgroup\$
    – clabacchio
    Mar 27, 2012 at 16:54
  • 1
    \$\begingroup\$ I don't work in audio, but isn't 80 dB (I assume you mean 10,000x voltage gain, since you don't have any load, and therefore you have no power gain) a lot of gain to be trying to get from a single op-amp stage? \$\endgroup\$
    – The Photon
    Mar 28, 2012 at 2:47

1 Answer 1

Reset to default
2
\$\begingroup\$

I finally found the right transfer function (thanks to slateraptor at eevblog): $$\frac{V_o}{V_i}=-(\frac{R_3}{R_1+R_3})(\frac{j\omega R_2C}{-\omega^2(\frac{R_1R_3}{R_1+R_3})C^2R_2+j\omega(\frac{R_1R_3}{R_1+R_3})2C+1}) $$

I actually just forgot a \$\frac{1}{R_1}\$ factor in my first function.

This produces a much more logical bode diagram where only the 1kHz frequency is amplified: bode diagram

\$\endgroup\$
1
  • \$\begingroup\$ Note that being an active filter, both transfer functions are plausible in an ideal world; you can offset your Bode diagram acting on the resistors \$\endgroup\$
    – clabacchio
    Mar 28, 2012 at 12:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.